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1999 = 1 (mod 9)

so 19991999

= 1 (mod 9)

Therefore we conclude that E=1.

2. There is a 50m long army platoon marching ahead. The last person in the platoon

wants to give a letter to the first person leading the platoon. So while the platoon is

marching he runs ahead, reaches the first person and hands over the letter to him

and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time. Assuming that he

ran the whole distance with uniform speed.

Answer

The last person covered 120.71 meters.

It is given that the platoon and the last person moved with uniform speed. Also, they both

moved for the identical amount of time. Hence, the ratio of the distance they covered - while

person moving forward and backword - are equal.

Let's assume that when the last person reached the first person, the platoon moved X meters

forward.

Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved

X meters.

Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon

moved (50-X) meters.

Now, as the ratios are equal,

(50+X)/X = X/(50-X)

(50+X)*(50-X) = X*X

Solving, X=35.355 meters

Thus, total distance covered by the last person

= (50+X) + X

= 2*X + 50

= 2*(35.355) + 50

= 120.71 meters

Note that at first glance, one might think that the total distance covered by the last person is

100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the

relative distance covered by the last person i.e. assuming that the platoon is stationary.

3. If you take a marker & start from a corner on a cube, what is the maximum

number of edges you can trace across if you never trace across the same edge

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twice, never remove the marker from the cube, & never trace anywhere on the

cube, except for the corners & edges?

Answer9

To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always

starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing

along the edges of a cube.

There is no need to start from other corners or edges of the cube, as you will only be repeating

the same combinations. The process is a little more involved than this, but is useful for solving

many types of spatial puzzles.

4. One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and

another is a Doctor.

If the Doctor is a male, then the Engineer is a male.

If the Engineer is younger than the Doctor, then the Engineer and the Doctor

are not blood relatives.

If the Engineer is a female, then she and the Doctor are blood relatives.

Can you tell who is the Doctor and the Engineer?

Answer : Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.

Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the

Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is

a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.

Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr.

Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.

Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his

son . It is not possible to determine anything further.

5. Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not

necessarily in the same order.

Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife

partners her husband and Cam does not play bridge.

Who is married to Cam?

Answer : Carrie is married to Cam.

"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."

It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.

As Cam does not play bridge, Billy's husband must be Laurie.Hence, Carrie is married to Cam.

6. There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many

as they had. After a month Y gave as many tractors to X and Z as many as they

have. After a month Z did the same thing. At the end of this transaction each one of

them had 24.

Find the tractors each originally had?

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Answer : One way to solve it is by making 3 equations and solve them simultaneously. But

there is rather easier way to solve it using Back tracing.

It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as

many as they had. It means that after getting tractors from Z their tractors got doubled. So

before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)

Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42

tractors i.e. (6, 42, 24)

Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39

tractors i.e. (39, 21, 12)

Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.

7. A certain street has 1000 buildings. A sign-maker is contracted to number the

houses from 1 to 1000. How many zeroes will he need?

Answer

The sign-maker will need 192 zeroes.

Divide 1000 building numbers into groups of 100 each as follow:

(1..100), (101..200), (201..300), ....... (901..1000)

For the first group, sign-maker will need 11 zeroes.

For group numbers 2 to 9, he will require 20 zeroes each.

And for group number 10, he will require 21 zeroes.

The total numbers of zeroes required are

= 11 + 8*20 + 21

= 11 + 160 + 21

= 192

8. There are 9 coins. Out of which one is odd one i.e weight is less or more. How

many iterations of weighing are required to find odd coin?

Answer

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier

or lighter.

1.Take 8 coins and weigh 4 against 4.

If both are not equal, goto step 2

If both are equal, goto step 3

2.One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1,

H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and

L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4,

X) where X is one coin remaining in intial weighing.

If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.

If both are equal, L4 is the odd coin and is lighter.

If L2 is light, L2 is the odd coin and is lighter.

If L3 is light, L3 is the odd coin and is lighter.

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If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight

H1 against H2

If both are equal, there is some error.

If H1 is heavy, H1 is the odd coin and is heavier.

If H2 is heavy, H2 is the odd coin and is heavier.

If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is

lighter. Weight H3 against H4

If both are equal, L1 is the odd coin and is lighter.

If H3 is heavy, H3 is the odd coin and is heavier.

If H4 is heavy, H4 is the odd coin and is heavier.

3.The remaining coin X is the odd one. Weigh X against the anyone coin used in initial

weighing.

If both are equal, there is some error.

If X is heavy, X is the odd coin and is heavier.

If X is light, X is the odd coin and is lighter.

9. In a sports contest there were m medals awarded on n successive days (n > 1).

1.On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.

2.On the second day 2 medals and 1/7 of the now remaining medals was awarded; and

so on.

3.On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded altogether?

Answer

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals

On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medalsOn day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals

On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals

On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals

On day 6: Medals awarded 6

10. A number of 9 digits has the following properties:

The number comprising the leftmost two digits is divisible by 2, that comprising the

leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and

so on for the nine digits of the number i.e. the number formed from the first n digits is

divisible by n, 2

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occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th

position will contain 5 as 0 does not